Given this functions series : $\sum_{n=0}^{\infty}x^n\sin (nx)$, I need to find the ratio where it converges. I don't see how can I change it into a form where I'll be able to use Cauchy-Hadmard or d'Alambert theorems into order to find R, the radius of convergence.

Rewrite it as the imaginary part of $\sum_{n=0}^\infty (xe^{ix})^n$, perhaps? Then it clearly converges for (real) $|x|<1$. But does divergence of the sum mean divergence of the imaginary part of the sum? Not sure.
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Thomas AndrewsFeb 7 '12 at 20:02

Lemma: If $\frac{x}\pi$ is irrational, then there are infinitely many positive integers $n$ such that $|\sin nx| > \frac{1}{2}$. If you can prove that, then for any $R>1$ you can find such an x in $(1,R)$, and the sequence $x\sin nx$ cannot converge to zero, and hence its sum cannot converge.
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Thomas AndrewsFeb 7 '12 at 20:36

@JavaMan: I'm not sure I understand your conclusion from these inequalities, can you extend it? Thnaks!
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JozefFeb 7 '12 at 20:59

@Jozef: All I am using is that if $\sum a_n \leq \sum b_n$, and $\sum b_n$ converges, then so does $\sum a_n$. This shows that the radius of convergence of the first sum is at least as large as the radius of convergence for the second sum.
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JavaManFeb 7 '12 at 21:08

1 Answer
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By comparison with $\sum |x|^n$, our series converges absolutely if $|x|<1$. Let's see why things go bad for most $|x|\ge 1$. Except when $x$ is of the form $k\pi$, the terms do not have limit $0$.

To do this, you will have to show that except in the case when $x$ is an integer multiple of $\pi$, we can find a positive $\alpha$ such that infinitely many integers $n$, $|\sin(nx)|>\alpha$. Hint: Suppose that by bad luck $\sin(nx)$ is awfully close to $0$. Show that $\sin((n+1)x)$ isn't.

Remark: I am not sure about the use of the term convergence radius. With power series, we have divergence if $|x|$ is greater than the convergence radius. Here we mostly have divergence, but the points $k\pi$ are exceptional.